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Evaluate $\int ydx + zdy + xdz$ where $C $ is intersection of $x+y=2$ and $x^2+y^2+z^2=2(x+y) $ traversed counterclockwise as viewed from origin

I am using Stokes' theorem to solve this question so

We want $\int \int curl F.N \; dS$ where $N$ is the normal unit vector to surface S, where S is a surface bounded by $C$

$F = \langle y,z,x\rangle$

$curl F = \langle -1,-1,-1 \rangle $

I take $S $ on the plane $x+y = 2$

$\nabla (x+y) = \langle 1,1,0 \rangle = A(say)$

Then unit normal vector $N = \langle -1/\sqrt2,-1/\sqrt2, 0\rangle $ {Multiplied by $-1$ because we are viewing it from origin }

$curl F.N= \sqrt(2)$

Now intersection of $x+y=2 $ and $x^2+y^2+z^2 = 2(x+y) $ gives

$x^2 + y^2 + z^2 = 4$

To get projection onto $xy$ plane $z=0$ we get

$x^2 +y^2 =4$

Now I am stuck in finding $dS$

How do I get dS = $\sqrt{z_x^2 +z_y^2 +1}dA$ where $A:x^2 +y^2 =4$

This is because my $S:x+y=2$ has no $z$ term

Christian Blatter
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So Lo
  • 1,567

1 Answers1

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Since $\nabla\times F\cdot \vec{N}=\sqrt{2}$, you only need to evaluate $$\sqrt{2}\iint_S dS$$ which is $\sqrt{2}$ multiplied by the area of the disk. (Note that the intersection is a disk, and it passes through the center of the original ball.) The original ball can be written as $$(x-1)^2+(y-1)^2+z^2=2$$ whose center is $(1,1,0)$. The plane $x+y=2$ also passes through the center $(1,1,0)$. Hence the area of the disk is $\pi r^2=2\pi$, and the answer for the integral is $2\sqrt{2}\pi$.

Alternatively, if we project this surface to $xz$-plane, (we cannot project it to $xy$-plane since it is perpendicular to the $xy$-plane) we substituting $y=2-x$ into the other equation to obtain $$x^2+(2-x)^2+z^2=4\implies (x-1)^2+\frac{z^2}{2}=1.$$ Now setting $y=2-x$, we have $$\sqrt{1+y_x^2+y_z^2}=\sqrt{2}.$$ Computing $$\sqrt{2}\cdot\sqrt{2}\iint_A dA$$ where $A$ is region enclosed by the ellipse, we obtain $2\sqrt{2}\pi$, same as above.

KittyL
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  • So if integrand in surface integral is $1$ then I can replace $dS$ with $dA$ where $A$ is the projection onto the $xy$ plane? – So Lo Dec 15 '17 at 14:26
  • Also answer is given at back is $2\sqrt{2} \pi$ – So Lo Dec 15 '17 at 14:28
  • @SoLo: The answer is correct since you need to multiply the area by $\sqrt{2}$. In this problem you cannot project it onto the $xy$ plane since it is perpendicular to the $xy$ plane. You can of course project it onto $xz$ or $yz$ plane. But a better way is just to use the area since the area is very easy to find. Remember that the goal of surface integral is to find the area of the surface. – KittyL Dec 15 '17 at 14:29
  • KittyL can you please tell me how can I project it onto one of the planes you mentioned. – So Lo Dec 15 '17 at 14:32
  • The intersection is $x^2 +y^2 +z^2=4$ so to project onto $yz$ I ll keep $x=0$ ie $y^2 + z^2=4$? – So Lo Dec 15 '17 at 14:34
  • You cannot say that the intersection is $x^2+y^2+z^2=4$. You can only say that it also gives the intersection. Keep in mind that the intersection is a 2 dimensional circle. I'll think about the other projection. – KittyL Dec 15 '17 at 14:42
  • @SoLo: I added the projection method. – KittyL Dec 15 '17 at 16:19
  • thank you so much.It is clear to me now – So Lo Dec 15 '17 at 18:25
  • @SoLo: You are welcome. Glad it helped. – KittyL Dec 15 '17 at 20:54
  • @KittyL if we project onto XZ Plane then radius isdifferent then when it is projected in Yx plane – J. Deff Jun 27 '19 at 07:50
  • @J.Deff: We cannot project onto the $xy$-plane since $S$ is perpendicular to it. – KittyL Jun 27 '19 at 18:41
  • @KittyL yes but If we project onto XZ AND YZ plane then radius of circle projected is different – J. Deff Jun 29 '19 at 08:58
  • @J.Deff: They are the same. You just need to substitute $x=2-y$. The equation will be similar, with $x$ replaced by $y$. – KittyL Jun 29 '19 at 10:38
  • HI,I found this exercise in apostol's calculus vol.2 in chapter 10 wich is about line integrals.I wonder if it can be solved without stokes theorem wich I dont know yet and without double integral i.e only with parametric equations.Thank you! – samer abdullah Jan 22 '21 at 08:04